a = 12
a[1] 12Set up eduroam
Go back and make sure you have the MOST RECENT version of R and RStudio
A single number is a scalar
\[a = 12\]
We already talked about vectors in R.
How would you define them?
A collection of concatenated elements
Column vector
\[ \overrightarrow b = \begin{bmatrix} 12 \\ 14 \\ 15 \end{bmatrix} \]
Row vector
\[ \overrightarrow c = \begin{bmatrix} 12 & 14 & 15 \end{bmatrix} \]
\[\overrightarrow d = \begin{bmatrix} 12 & 7 & -2 & 3 & -1 \end{bmatrix}\]
\[\sum_{i=1}^3 d_i\]
\[\overrightarrow d = \begin{bmatrix} 12 & 7 & -2 & 3 & -1 \end{bmatrix}\]
\[\sum_{i=1}^3 d_i = 12+7+(-2)\]
\[\overrightarrow d = \begin{bmatrix} 12 & 7 & -2 & 3 & -1 \end{bmatrix}\]
\[ \sum_{i=1}^n d_i \]
\[\overrightarrow d = \begin{bmatrix} 12 & 7 & -2 & 3 & -1 \end{bmatrix}\]
\[ \sum_{i=1}^n d_i = 12 + 7 + (-2) + 3 + (-1) \]
\[\overrightarrow d = \begin{bmatrix} 12 & 7 & -2 & 3 & 1 \end{bmatrix}\]
\[\prod_{i=1}^n d_i\]
\[\overrightarrow d = \begin{bmatrix} 12 & 7 & -2 & 3 & 1 \end{bmatrix}\]
\[\prod_{i=1}^n d_i = 12 \cdot 7 \cdot (-2) \cdot 3 \cdot (-1)\]
\[\overrightarrow d = \begin{bmatrix} 12 & 7 & -2 & 3 & 1 \end{bmatrix}\]
Practice: Write and calculate the product of the first three elements of \(\overrightarrow{d}\)
\[A = \begin{bmatrix} 12 & 14 & 15 \\ 115 & 22 & 127 \\ 193 & 29 & 219 \end{bmatrix}\]
Why are they important?
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[4,] 10 11 12
\[ B= \begin{bmatrix} b_{11} & b_{12} & b_{13} & \ldots & b_{1n} \\ b_{21} & b_{22} & b_{23} & \ldots & b_{2n} \\ \vdots & \vdots & \vdots & \ldots & \vdots \\ b_{m1} & b_{m2} & b_{m3} & \ldots & b_{mn} \end{bmatrix} \]
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[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12 [,1] [,2] [,3]
[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12 [,1] [,2] [,3]
[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12Practice: Write code to find the index corresponding to number 12
\[A \pm B=C\]
\[c_{ij}=a_{ij} \pm b_{ij} \text{ }\forall i,j\]
Add or subtract each corresponding element
Need \(A\) and \(B\) to have exactly the same dimensions
\[\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix} \pm \begin{bmatrix} b_{11} & b_{12} & b_{13}\\ b_{21} & b_{22} & b_{23}\\ b_{31} & b_{32} & b_{33} \end{bmatrix}\] \[=\] \[\begin{bmatrix} a_{11}\pm b_{11} & a_{12}\pm b_{12} & a_{13}\pm b_{13}\\ a_{21}\pm b_{21} & a_{22}\pm b_{22} & a_{23}\pm b_{23}\\ a_{31}\pm b_{31} & a_{32}\pm b_{32} & a_{33}\pm b_{33} \end{bmatrix}\]
\[A= \begin{bmatrix} 3 & -1 & 2 \\ 9 & 4 & 6 \end{bmatrix}\]
\[B= \begin{bmatrix} 5 & 2 & 0 \\ 9 & 3 & 4 \end{bmatrix}\]
\[A= \begin{bmatrix} 3 & -1 & 2 \\ 9 & 4 & 6 \end{bmatrix}\]
\[A= \begin{bmatrix} 3 & -1 & 2 \\ 9 & 4 & 6 \end{bmatrix}\]
Do one with code and the other by hand.
1) Calculate \(A + B\)
\[A= \begin{bmatrix} 1 & 0 \\ -2 & -1 \end{bmatrix}\]
\[B = \begin{bmatrix}
5 & 1 \\
2 & -1
\end{bmatrix}\]
2) Calculate \(A - B\)
\[A= \begin{bmatrix} 6 & -2 & 8 & 12 \\ 4 & 42 & 8 & -6 \end{bmatrix}\]
\[B = \begin{bmatrix} 18 & 42 & 3 & 7 \\ 0 & -42 & 15 & 4 \end{bmatrix}\]
\[A = \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}\] \[cA = \begin{bmatrix} ca_{11} & ca_{12} & ca_{13}\\ ca_{21} & ca_{22} & ca_{23}\\ ca_{31} & ca_{32} & ca_{33} \end{bmatrix}\]
Calculate \(2\times A\) and \(-3 \times B\).
\[A= \begin{bmatrix} 1 & 4 & 8 \\ 0 & -1 & 3 \end{bmatrix}\] \[ B = \begin{bmatrix} -15 & 1 & 5 \\ 2 & -42 & 0 \\ 7 & 1 & 6 \end{bmatrix}\]
Do one by hand, one in code
Matrices must be conformable
For \(A \times B\) we need \(A_{i \times k}\) and \(B_{k \times j}\)
The result will have \(i \times j\) dimensions
Order matters! \(A \times B \neq B \times A\) even if both operations are conformable
Multiply each row by each column, summing up each pair of multiplied terms (dot product)
The element in position \(ij\) is the sum of the products of elements in the \(i\)th row of the first matrix (\(A\)) and the corresponding elements in the \(j\)th column of the second matrix (\(B\)). \[c_{ij}=\sum_{k=1}^n a_{ik}b_{kj}\]
Furniture company construction costs ($)
| Chair | Sofa | |
|---|---|---|
| Wood | 100 | 150 |
| Cloth | 270 | 420 |
| Feathers | 130 | 195 |
\[C = \begin{bmatrix} 100 & 150\\ 270 & 420\\ 130 & 195 \end{bmatrix}\]
Monthly production
| Product | Quantity |
|---|---|
| Chair | 45 |
| Sofa | 30 |
\[Q = \begin{bmatrix} 45 \\ 30 \end{bmatrix}\]
Total expenditure
\[ E = CQ = \begin{bmatrix} 100 & 150\\ 270 & 420\\ 130 & 195 \end{bmatrix} \begin{bmatrix} 45 \\ 30 \end{bmatrix}\]
| Chair | Sofa | |
|---|---|---|
| Wood | 100 | 150 |
| Cloth | 270 | 420 |
| Feathers | 130 | 195 |
| Product | Quantity |
|---|---|
| Chair | 45 |
| Sofa | 30 |
Total expenditure
\[ E = CQ = \begin{bmatrix} 100 & 150\\ 270 & 420\\ 130 & 195 \end{bmatrix} \begin{bmatrix} 45 \\ 30 \end{bmatrix}\]
\[ = \begin{bmatrix} (100)(45) + (150)(30) \\ (270)(45) + (420)(30) \\ (130)(45) + (195)(30) \end{bmatrix} = \begin{bmatrix} 9,000 \\ 24,750 \\ 11,700 \end{bmatrix} \]
Check in R with the %*% operator to see if these are conformable
\[ \begin{aligned} B_{4 \times 1}= \begin{bmatrix} 2 \\ 3\\ 4\\ 1 \end{bmatrix} M_{3 \times 3} = \begin{bmatrix} 1 & 0 & 2\\ 1 & 2 & 4\\ 2 & 3 & 2 \end{bmatrix} L_{2 \times 3} = \begin{bmatrix} 6 & 5 & -1\\ 1 & 4 & 3 \end{bmatrix} \end{aligned} \]
Addition and subtraction:
Multiplication:
\(AB \neq BA\)
If \(AB = BA\) we say they commute
\(A(BC) = (AB)C\)
\(A(B+C) = AB + AC\)
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \]
\(m=n\)
Diagonal \(d(A)=\{1,5,9\}\)
Trace \(tr(A) = 1+5+9 = 15\)
\[ D = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 6 \end{bmatrix} \]
\[ S = \begin{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end{bmatrix} \]
\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
\[J = \begin{bmatrix} 4 & 5\\ 3 & 0\\ 7 & -2 \end{bmatrix} \quad J' = J^T = \begin{bmatrix} 4 & 3 & 7 \\ 5 & 0 & -2 \end{bmatrix}\]
\[A × A^{-1} = I\]
The inverse of \(A\) is \(A^{-1}\) iff
\[AA^{-1} = A^{-1}A = I\]
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[3,] 5 2 4\[\widehat{\beta} = \underbrace{(X'X)^{-1}X'Y}_\text{"X prime X inverse, X prime Y"}\]